Solve $\int_T \frac{1}{x^2+y^2}dxdy$ I have to solve the integral of the function:
$$f(x,y)=\frac{1}{x^2+y^2}$$ 
on this set: $T=\{(x,y): y^2\lt x\lt y\}$
I think to substitute with polar coordinates, but i can't find the integration limits.
 A: For all $(x,y) \in T$ you have $y^2 \le y$, so necessarily $y \in [0,1]$. As a consequence
$$T= \{ (x,y) : y \in [0,1]  , x \in [y^2, y] \}$$
Now, your integral is
$$\int_0^1 \int_{y^2}^y \frac{1}{x^2+y^2} \mathrm{d}x \ \mathrm{d} y = \int_0^1 \left[ \frac{1}{y} \arctan \left( \frac{x}{y} \right)\right]_{x=y^2}^{x=y} \mathrm{d} y = \int_0^1 \frac{1}{y} \left( \arctan 1 - \arctan y\right) \mathrm{d}y$$
which is a divergent integral.
A: Make a drawing, and get convinced that either
$$\color{red}{0\le x\le 1\;,\;\;x\le y\le \sqrt x}\;,\;\;\text{or}\;\;\color{green}{0\le y\le1\;,\;\;y^2\le x\le y}$$
Using now polar coordinates you want 
$$\frac\pi4\le\theta\le\frac\pi2\;,\;\;\text{since}\;\;r\cos\theta\le r\sin\theta\le\sqrt{r\cos\theta}$$
and the first inequality gives
$$\tan\theta\ge1\implies \theta\ge\frac\pi 4\;,\;\;\text{whereas the second one gives}\;\;$$
$$r^2\cos^2\theta\le r\cos\theta\implies r\le\frac1{\cos\theta}$$
We get a divergent integral, and this can also be clearly seen (imo) in your original one...Are you sure of the integration domain?
